Question

The value of $\int_{2}^{4} (x-2)(x-3)(x-4)\, dx$ is equal to:

• A. $\frac{1}{2}$
• B. $2$
• C. $3$
• D. $\frac{1}{3}$
• E. $0$

Hint:

If roots are symmetric around midpoint, the cubic becomes odd → integral becomes zero instantly.

Answer 1

Answer: (E)

Concept: If a function becomes odd after shifting to the midpoint of the interval, then: \[ \int_{-a}^{a} f(x)\,dx = 0 \]

Step 1: Shift origin to midpoint.
Given: \[ f(x) = (x-2)(x-3)(x-4) \] Midpoint of interval \( [2,4] \) is \( x = 3 \). Let \( x = t + 3 \)

Step 2: Transform the function.
\[ f(t+3) = (t+1)(t)(t-1) \] \[ = t(t^2 - 1) = t^3 - t \]

Step 3: Check symmetry.
\[ g(t) = t^3 - t \] \[ g(-t) = -g(t) \] So, function is odd.

Step 4: Change limits.
\[ x = 2 \Rightarrow t = -1, \quad x = 4 \Rightarrow t = 1 \]

Step 5: Evaluate integral.
\[ \int_{2}^{4} (x-2)(x-3)(x-4)\,dx = \int_{-1}^{1} (t^3 - t)\,dt = 0 \]

Step 6: Final answer.
\[ \boxed{0} \]